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# Find the HCF of the following numbers :

(a) 18, 48

(b) 30, 42

(c) 18, 60

(d) 27, 63

(e) 36, 84

(f) 34, 102

(g) 70, 105, 175

(h) 91, 112, 49

(i) 18, 54, 81

(j) 12, 45, 75

**Solution:**

We will use the concept of the HCF(Highest Common Factor) to solve this.

(a) 18, 48

The prime factorizations of 18 and 48 are as follows:

18 = 2 × 3 × 3

48 = 2 × 2 × 2 × 2 × 3

The common factors, in this case, are 2 and 3.

HCF = 2 × 3 = 6

Hence, the HCF of 18, 48 is 6

(b) 30, 42

The prime factorizations of 30 and 42 are as follows:

30 = 2 × 3 × 5

42 = 2 × 3 × 7

The common factors, in this case, are 2 and 3.

HCF = 2 × 3 = 6

Hence, the HCF of 30, 42 is 6.

(c) 18, 60

The prime factorizations of 18 and 60 are as follows:

18 = 2 × 3 × 3

60 = 2 × 2 × 3 × 5

The common factors, in this case, are 2 and 3.

HCF = 2 × 3 = 6

Hence, the HCF of 18, 60 is 6.

(d) 27, 63

The prime factorizations of 27 and 63 are as follows:

27 = 3 × 3 × 3

63 = 3 × 3 × 7

The common factors, in this case, are 3 and 3.

HCF = 3 × 3 = 9

Hence, the HCF of 27, 63 is 9

(e) 36, 84

The prime factorizations of 36 and 84 are as follows:

36 = 2 × 2 × 3 × 3

84 = 2 × 2 × 3 × 7

The common factors, in this case, are 2, 2, and 3.

HCF = 2 × 2 × 3 = 12

Hence, the HCF of 36, 84 is 12

(f) 34, 102

The prime factorizations of 34 and 102 are as follows:

34 = 2 × 17

102 = 2 × 3 × 17

The common factors, in this case, are 2 and 17.

HCF = 2 × 17 = 34

Hence, the HCF of 34, 102 is 34

(g) 70, 105, 175

The prime factorizations of 70, 105, and 175 are as follows:

70 = 2 × 5 × 7

105 = 3 × 5 × 7

175 = 5 × 5 × 7

The common factors, in this case, are 5 and 7.

HCF = 5 × 7 = 35

Hence, the HCF of 70, 105, 175 is 35

(h) 91, 112, 49

The prime factorizations of 91, 112, and 49 are as follows:

91 = 7 × 13

112 = 2 × 2 × 2 × 2 × 7

49 = 7 × 7

The common factors, in this case, are 7.

HCF = 7

Hence, the HCF of 91, 112, 49 is 7

(i) 18, 54, 81

The prime factorizations of 18, 54, and 81 are as follows:

18 = 2 × 3 × 3

54 = 2 × 3 × 3 × 3

81 = 3 × 3 × 3 × 3

The common factors, in this case, are 3 and 3.

HCF = 3 × 3 = 9

Hence, the HCF of 18, 54, 81 is 9

(j) 12, 45, 75

The prime factorizations of 12, 45 and 75 are as follows:

12 = 2 × 2 × 3

45 = 3 × 3 × 5

75 = 3 × 5 × 5

The common factors, in this case, are 3.

HCF = 3

Hence, the HCF of 12, 45, 75 is 3

You can also use the HCF Calculator to answer this.

NCERT Solutions for Class 6 Maths Chapter 3 Exercise 3.6 Question 1

## Find the HCF of the following numbers : (a) 18, 48 (b) 30, 42 (c) 18, 60 (d) 27, 63 (e) 36, 84 (f) 34, 102 (g) 70, 105, 175 (h) 91, 112, 49 (i) 18, 54, 81 (j) 12, 45, 75

**Summary:**

The HCF of the given numbers are a) 6, b) 6, c) 6, d) 9, e) 12, f) 34, g) 35, h) 7, i) 9, j) 3.

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